Difference between revisions of "2017 AMC 10A Problems/Problem 4"

(Solution 1)
(Solution 1)
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How did you choose to do <math>27\times30?</math>
How did you choose to do <math>27\times30?</math>
~A question by greenstarwins
~A question by greenstar
==Solution 2==
==Solution 2==

Revision as of 12:46, 3 February 2021


Mia is "helping" her mom pick up $30$ toys that are strewn on the floor. Mia’s mom manages to put $3$ toys into the toy box every $30$ seconds, but each time immediately after those $30$ seconds have elapsed, Mia takes $2$ toys out of the box. How much time, in minutes, will it take Mia and her mom to put all $30$ toys into the box for the first time?

$\textbf{(A)}\ 13.5\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 14.5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 15.5$

Solution 1

Every $30$ seconds, $3-2=1$ toys are put in the box, so after $27\cdot30$ seconds, there will be $27$ toys in the box. Mia's mom will then put $3$ toys into the box, and we have our total amount of time to be $27\cdot30+30=840$ seconds, which equals $14$ minutes. $\boxed{(\textbf{B})\ 14}$

How did you choose to do $27\times30?$ ~A question by greenstar

Solution 2

Though Mia's mom places $3$ toys every $30$ seconds, Mia takes out $2$ toys right after. Therefore, after $30$ seconds, the two have collectively placed $1$ toy into the box. Therefore by $13.5$ minutes, the two would have placed $27$ toys into the box. Therefore, at $14$ minutes, the two would have placed $30$ toys into the box. Though Mia may take $2$ toys out right after, the number of toys in the box first reaches $30$ by $14$ minutes. $\boxed{(\textbf{B})\ 14}$

Video Solution




See also

2017 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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